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Newtons 3rd Law (Ropes, Pulley)

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A 2.5 kg mass is attached to a 3.5 kg mass by a light string that passes over a friction-less pulley. The masses are released.

    a) What is the magnitude of the acceleration of the masses?
    b) What is the tension in the string?

    2. Relevant equations

    (I think) Ft - mg = ma

    3. The attempt at a solution

    I drew 2 free body diagrams for both the weights and they both have Fg and Ft on the y.

    M5wh7.jpg

    On this 6th line, starting with (2.5)(a)... I made it so that Ft2 = Ft2
    My answer was suppose to be 6.54 m/s in the work I showed because the acceleration of the bigger weight will accelerate negatively. So ignore the minus sign for my answer to a =
    I think I've done the best I could, but my answer is still wrong, not sure why.
     
    Last edited: Oct 30, 2012
  2. jcsd
  3. Oct 30, 2012 #2

    lewando

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    The magnitude of the acceleration will be the same, but in opposite directions.
     
  4. Oct 30, 2012 #3
    Yeah, I knew that, I should have included that too. I need actual with with the question though.
     
  5. Oct 30, 2012 #4

    lewando

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    You have 2 equations, 2 unknowns (one needs fixing-- one of the masses should be accelerating upwards-- use "a", the other, downwards-- use "-a"). Solve for a.
     
  6. Oct 30, 2012 #5
    I've done some changes, but I am still getting it wrong...
     
  7. Oct 30, 2012 #6

    lewando

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    Since the heavier one is accelerating downward, replace "a" in line 2 with "-a". Try again.
     
  8. Oct 30, 2012 #7
    Ohhhhhhh, thank youuuuuuuuuuuuu. I re-did my whole thing and now it works!!!
     
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